String Functions

This is a discussion on String Functions within the C++ Programming forums, part of the General Programming Boards category; I'm trying to write some code that will allow me to read in a character array from the user. I ...

  1. #1
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    Smile String Functions

    I'm trying to write some code that will allow me to read in a character array from the user. I managed to get something that will read everything (up to the specified length), including spaces. Since the code for that one function is slightly long, I tried to put it into a function. My problem is that the code inside the function doesn't read anything after the first space is put in.

    Here's the working code that I have:

    *********************************************
    #include <stdio.h>

    int main()
    {
    char cString[15] = "\0";

    printf("Please type a name:");

    int nCurrChar = 0;
    for(int i = 0; (i<sizeof(cString)) && ((nCurrChar = getchar()) !=EOF) && (nCurrChar != '\n'); i++)
    cString[i] = (char) nCurrChar;
    cString[i] = '\0';

    printf("%s\n", cString);

    return 0;
    }
    *********************************************

    Here's the code that I have that isn't working:

    *********************************************
    #include <stdio.h>
    #include <string.h>

    void ReadString(char string[]);

    int main()
    {
    printf("Please type a name:");

    char cName[15] = "\0";

    ReadString(cName);

    printf("%s", cName);

    return 0;
    }

    void ReadString(char cString[])
    {
    int nCurrChar = 0;
    for(int i = 0; (i<sizeof(cString)) && ((nCurrChar = getchar()) !=EOF) && (nCurrChar != '\n'); i++)
    cString[i] = (char) nCurrChar;
    cString[i] = '\0';
    }

    *********************************************

    I copied and pasted the code that I put into the function directly from what I had above it, so I'm not sure what's going on. Any help would be greatly appreciated. :-)

  2. #2
    Ethereal Raccoon Procyon's Avatar
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    The problem isn't spaces; the problem is your use of the sizeof operator. This doesn't work on C-arrays like you might expect. A C-array is actually just a pointer, so using sizeof on it will return the size in bytes of the pointer itself, not the array it indexes. No matter your array length, sizeof of a char* will be 4 bytes, invariably tripping the first for-loop exit condition and ending input after four characters.

    You can try using the C++ string class instead, or make your own string class such as this:

    Code:
    class mystring
    {
      private:
      char *cstring;
      unsigned int length;
    
      public:
      mystring(unsigned int initlength);
      ~mystring();
      unsigned int size()  {return length;};
      char *getstring(); {return cstring;};
      void readstring();
    };
    The implementation of readstring() would be almost the same as your ReadString, but with size() replacing sizeof(). Of course, you'd have to know about constructors, destructors, and dynamic memory to do this.
    Last edited by Procyon; 09-01-2001 at 12:29 PM.

  3. #3
    zen
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    ..or if you want to continue to use c-strings you could pass the size of the array into the function as a parameter.

  4. #4
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    would strlen() work?
    int strlen(char* sz)
    Requires that the string is terminated with the null char and returns the length as an integer...
    zMan

  5. #5
    Ethereal Raccoon Procyon's Avatar
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    No, because the strlen() of an uninitialized array will be potentially unlimited, and for an array initialized to "\0" it will be zero.

    I think zen's strategy is the simplest.

  6. #6
    and the hat of wrongness Salem's Avatar
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    > the problem is your use of the sizeof operator. This doesn't work on C-arrays like you might expect.
    This isn't exactly true - it depends on the context of the 'array' in question as to whether it 'works' or not.

    In the first case, the array is a proper array and sizeof will give you the answer you expect. sizeof will tell you the size of an array.

    In the second case, the cString isn't an array (despite the use of [] in the declaration), it is simply a pointer (all arrays are passed as pointers to the first element). In this case, sizeof will only tell you the size of that pointer (which is 4 in most of common environments).

    > ..or if you want to continue to use c-strings you could pass the size of the array into the function as a parameter.

    Code:
    #include <stdio.h>
    #include <string.h>
    void ReadString(char string[], int size);
    int main()
    {
        printf("Please type a name:");
        char cName[15] = "\0";
        ReadString(cName,sizeof(cName));
        printf("%s", cName);
        return 0;
    }
    void ReadString(char cString[], int size)
    {
        int nCurrChar = 0;
        for(int i = 0; (i<size) && ((nCurrChar = getchar()) !=EOF) && (nCurrChar != '\n'); i++)
            cString[i] = (char) nCurrChar;
        cString[i] = '\0';
    }
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
    If at first you don't succeed, try writing your phone number on the exam paper.
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  7. #7
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    Smile Thanks much!

    Thanks much guys! I really appreciate the help. :-)

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